Question: Simplify; express your answer in exponential form. Assume $r\neq 0, y\neq 0$. $\dfrac{{(r^{-2})^{5}}}{{(r^{-3}y)^{-3}}}$
To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-2}}$ to the exponent ${5}$ . Now ${-2 \times 5 = -10}$ , so ${(r^{-2})^{5} = r^{-10}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-3}y)^{-3} = (r^{-3})^{-3}(y)^{-3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-2})^{5}}}{{(r^{-3}y)^{-3}}} = \dfrac{{r^{-10}}}{{r^{9}y^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-10}}}{{r^{9}y^{-3}}} = \dfrac{{r^{-10}}}{{r^{9}}} \cdot \dfrac{{1}}{{y^{-3}}} = r^{{-10} - {9}} \cdot y^{- {(-3)}} = r^{-19}y^{3}$.